388. Longest Absolute File Path

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Suppose we abstract our file system by a string in the following manner:

The string “dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext” represents:

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dir
    subdir1
    subdir2
        file.ext

The directory dir contains an empty sub-directory subdir1 and a sub-directory subdir2 containing a file file.ext.

The string “dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext” represents:

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dir
    subdir1
        file1.ext
        subsubdir1
    subdir2
        subsubdir2
            file2.ext

The directory dir contains two sub-directories subdir1 and subdir2. subdir1 contains a file file1.ext and an empty second-level sub-directory subsubdir1. subdir2 contains a second-level sub-directory subsubdir2 containing a file file2.ext.

We are interested in finding the longest (number of characters) absolute path to a file within our file system. For example, in the second example above, the longest absolute path is “dir/subdir2/subsubdir2/file2.ext”, and its length is 32 (not including the double quotes).

Given a string representing the file system in the above format, return the length of the longest absolute path to file in the abstracted file system. If there is no file in the system, return 0.

Note:
The name of a file contains at least a . and an extension.
The name of a directory or sub-directory will not contain a ..
Time complexity required: O(n) where n is the size of the input string.

Notice that a/aa/aaa/file1.txt is not the longest file path, if there is another path aaaaaaaaaaaaaaaaaaaaa/sth.png.

解法1： HashMap

基本的思路是先把string按照\n分割，然后按照题意，对于每一个子字符串，他的\t的数量就代表他的层数。
这样的话可以从左往右扫描一遍，遇到一个新的level就存下当前的路径长度，不过如果当前是文件的话则从map中读取当前的level长度然后加上文件名长度就可以了。
每次存level的时候实际可以存上真实level+1，代表的是如果下一个文件的level是x，那么直接从map中读取x就是对应的path长度。
初始化的时候要放入map.put(0,0)

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class Solution {
    public int lengthLongestPath(String input) {
        int res = 0;
        Map<Integer, Integer> map = new HashMap<>();    // record the length at each level
        map.put(0, 0);
        for (String element : input.split("\n")) {
            int level = element.lastIndexOf("\t") + 1;
            element = element.substring(level);
            if (element.contains(".")) {
                res = Math.max(res, map.get(level) + element.length());
            } else {
                map.put(level + 1, map.get(level) + element.length() + 1);
            }
        }
        return res;
    }
}